\(\dfrac{1}{4}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{40}\) + \(\dfrac{1}{60}\) + \(\dfrac{1}{84}\) + \(x\) = 1

\(\dfrac{1}{2.2}\) + \(\dfrac{1}{2.6}\)+\(\dfrac{1}{2.12}\)+\(\dfrac{1}{2.20}\) + \(\dfrac{1}{2.30}\) + \(\dfrac{1}{2.42}\) +  \(x\) =1

\(\dfrac{1}{2}\).(\(\dfrac{1}{2}\) + \(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\) + \(\dfrac{1}{30}\)+ \(\dfrac{1}{42}\)) + \(x\) = 1

\(\dfrac{1}{2}\).( \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\)  + \(\dfrac{1}{4.5}\)+ \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)) + \(x\) = 1

\(\dfrac{1}{2}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)) + \(x\) = 1

\(\dfrac{1}{2}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{7}\)) + \(x\) = 1

\(\dfrac{1}{2}\).\(\dfrac{6}{7}\) + \(x\) = 1

       \(\dfrac{3}{7}\) + \(x\) = 1 

              \(x\) = 1 -  \(\dfrac{3}{7}\)

               \(x\) = \(\dfrac{4}{7}\) 

2A=1/2+1/6+1/12+1/20+1/30+1/42

    =1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7

    =1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7

    =1-1/7

    =6/7

\(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{40}+\dfrac{1}{60}+\dfrac{1}{84}\\ =\dfrac{2}{8}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{80}+\dfrac{2}{120}+\dfrac{2}{168}\\ =\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}+\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}\\ =\dfrac{1}{2}-\dfrac{1}{14}\\ =\dfrac{6}{14}\\ =\dfrac{3}{7}\)

\(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{40}+\dfrac{1}{60}+\dfrac{1}{84}\\ =\dfrac{2}{8}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{80}+\dfrac{2}{120}+\dfrac{2}{168}\\ =\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}+\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}\\ =\\ 1-\dfrac{1}{14}\\ =\dfrac{14}{14}-\dfrac{1}{14}\\ =\dfrac{13}{14}\)

trần ngọc nhi

Câu hỏi của Baekhyun - Toán lớp 6 - Học toán với OnlineMath

A = 1/4 + ... +1/84

A = 2/8 + 2/24 + ... + 2/168

A = 2/2.4 + 2/4.6 + ... + 2/12.14

A = 1/2 - 1/4 + 1/4 - 1/6 + .. + 1/12 - 1/14

A = 1/2 - 1/14

A = 6/14 = 3/7

A = 2/8 + 2/24 + ... + 2/168

A = 2/2.4 + 2/4.6 + ... + 2/12.14

A = 1/2 - 1/4 + 1/4 - 1/6 + .. + 1/12 - 1/14

A = 1/2 - 1/14

A = 3/7

\(A=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+...+\frac{2}{168}\)

\(A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{12.14}\)

\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{12}-\frac{1}{14}\)

\(A=\frac{1}{2}-\frac{1}{14}\)

\(A=\frac{3}{7}\)

ta nhóm lại với nhau

(\(\dfrac{1}{4}\)+\(\dfrac{1}{12}\))+(\(\dfrac{1}{24}\)+\(\dfrac{1}{40}\))+(\(\dfrac{1}{60}\)+\(\dfrac{1}{84}\))

=\(\dfrac{1}{3}\)+\(\dfrac{1}{15}\)+\(\dfrac{1}{35}\)

=\(\dfrac{3}{7}\)

đúng thì mình xin cái theo dõi nha bạn

3/7

tick đi đúng đó

Cho tg tren la A

A=\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)

\(A=2\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{14}\right)\)

\(A=2.\frac{3}{7}\)

\(A=\frac{6}{7}\)

Ta co : 

\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)

\(=2\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{14}\right)\)

\(=2.\frac{3}{7}\)

\(=\frac{6}{7}\)